\begin{answer}{constructzerocorrelation}
This is the same as question \ref{q:iszerocorrelationindependent}.
Here is the proof again, with the symbols updated to match the current question.
Determine the distribution of $Z$,
\begin{align*}
F_Z(x)  &= P(Z<x)  \\
&=
P( XY<x| Y = 1)P(Y=1) +
P( XY<x| Y = -1)P(Y=-1) \\
&=
P( X<x)(0.5) + P(-X<x)(0.5) \\
&=  (0.5)( P( X<x) + P(X \geq -x)) \\
&=  (0.5)( P( X<x) + P(X < x)) \quad \text{symmetry} \\
&=   P(X < x) \\
&=   F_X(x) \\
&=   \Phi(x)
\text{,}
\end{align*}
which means $Z \sim \text{Normal}(0, 1)$.

Intuitively, this makes sense.
You have a normal distribution multiplied by a variable that takes -1 and 1 with equal probability,
so you are drawing from a normal distribution, and then randomly mirroring the value in the y-axis.
Since the Normal$(0,1)$ distribution is symmetric about zero, this operation should not affect its distribution and thus the resulting distribution is also Normal$(0,1)$.


\end{answer}
